question_answer

Consider a sphere of radius R which carries a uniform charge density\[\rho \]. If a sphere of radius \[\frac{R}{2}\]is carved out of it, as shown, the ratio \[\frac{\left| {{{\vec{E}}}_{A}} \right|}{\left| {{{\vec{E}}}_{B}} \right|}\] of magnitude of electric field \[{{\vec{E}}_{A}}\] and \[{{\vec{E}}_{B}}\]respectively, at points A and B due to the remaining portion is: [JEE MAIN Held on 09-01-2020 Morning]

A) \[\frac{18}{34}\]           

B)        \[\frac{18}{54}\]

C) \[\frac{21}{34}\]                       

D) \[\frac{9}{17}\]

Correct Answer: D

Solution :

[d] \[{{E}_{A}}=\frac{\sigma \left( R/2 \right)}{3{{\varepsilon }_{0}}}=\left( \frac{\sigma R}{6{{\varepsilon }_{0}}} \right)\] \[{{E}_{B}}=\frac{\sigma R}{3{{\varepsilon }_{0}}}-\left( \frac{1}{4\pi {{\varepsilon }_{0}}} \right)\frac{\left( \sigma  \right)}{{{\left( \frac{3R}{2} \right)}^{2}}}\frac{4\pi }{3}{{\left( \frac{R}{2} \right)}^{3}}\] \[=\frac{\sigma R}{3{{\varepsilon }_{0}}}-\frac{\sigma R}{54{{\varepsilon }_{0}}}\] \[\Rightarrow {{E}_{B}}=\frac{17}{54}\left( \frac{\sigma R}{{{\varepsilon }_{0}}} \right)\] \[\left| \frac{{{E}_{A}}}{{{E}_{B}}} \right|=\frac{1\times 54}{6\times 17}=\left( \frac{9}{17} \right)\]